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Let \(S\) be the collection of vectors\(\left[\begin{array}{l}x \\\ y\end{array}\right]\) in \(\mathbb{R}^{2}\) that satisfy the given property. In each case, either prove that S forms a subspace of \(\mathbb{R}^{2}\) or give a counterexample to show that it does not. $$x y \geq 0$$

Short Answer

Expert verified
The set \( S \) is not a subspace of \( \mathbb{R}^2 \).

Step by step solution

01

Understanding the Problem

We need to determine if the set \( S \), consisting of vectors \( \begin{bmatrix} x \ y \end{bmatrix} \in \mathbb{R}^2 \) where the product \( xy \geq 0 \), forms a subspace of \( \mathbb{R}^2 \). To do this, we'll need to check the conditions necessary for a subset of \( \mathbb{R}^2 \) to be a subspace.
02

Checking the Zero Vector

A subspace must contain the zero vector. Check if \( S \) includes the zero vector \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \). Substituting, we find \( 0 \cdot 0 = 0 \geq 0 \), so the zero vector is in \( S \).
03

Closure Under Addition

A subspace must be closed under addition. Consider \( \begin{bmatrix} x_1 \ y_1 \end{bmatrix} \) and \( \begin{bmatrix} x_2 \ y_2 \end{bmatrix} \) in \( S \), meaning \( x_1y_1 \geq 0 \) and \( x_2y_2 \geq 0 \). Check if \( \begin{bmatrix} x_1 + x_2 \ y_1 + y_2 \end{bmatrix} \) belongs to \( S \). However, \((x_1 + x_2)(y_1 + y_2)\) does not necessarily satisfy \(\geq 0\), showing \( S \) is not closed under addition. E.g., vectors \( \begin{bmatrix} 1 \ 1 \end{bmatrix} \) and \( \begin{bmatrix} -1 \ -1 \end{bmatrix} \) belong to \( S \) but their sum \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) only incidentally satisfies the property, not due to closure.
04

Conclusion on Subspace Status

Given that \( S \) is not closed under addition, \( S \) does not satisfy the conditions to be a subspace of \( \mathbb{R}^2 \). Therefore, a counterexample shows that \( S \) is not a subspace of \( \mathbb{R}^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Closure Under Addition
The concept of closure under addition is crucial when determining whether a set of vectors forms a subspace. For a subspace, if you take any two vectors within the set, their sum must also be in that same set.
This ensures consistency and structure that mimic the vector space from which it is derived.
Consider vectors \( \begin{bmatrix} x_1 \ y_1 \end{bmatrix} \) and \( \begin{bmatrix} x_2 \ y_2 \end{bmatrix} \) in a subspace subset \( S \) of \( \mathbb{R}^2 \).
  • For closure, \( \begin{bmatrix} x_1 + x_2 \ y_1 + y_2 \end{bmatrix} \) must also be part of \( S \).
  • It's not just the individual sums of components that matter, but that their resulting combination satisfies any condition imposed on \( S \).
In the provided exercise, checking for closure under addition reveals a key failure for \( S \).
While individual conditions \( x_1 y_1 \geq 0 \) and \( x_2 y_2 \geq 0 \) may hold, the multiplication condition on sums \((x_1 + x_2)(y_1 + y_2) \) is not guaranteed to satisfy \( \geq 0 \).
This demonstrates non-closure, making \( S \) unsuitable as a subspace.
Zero Vector in Subspace
The zero vector, denoted \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \), plays a vital role in subspaces. Each subspace must include this zero vector to maintain the properties of a vector space.
This inclusion is essentially non-negotiable because:
  • The zero vector acts as the neutral element in vector addition, maintaining the identity for operations within the subspace.
  • Without it, a set cannot consistently adhere to vector space axioms, such as the existence of additive inverses and identities.
In the given problem, confirming \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \) satisfies \( 0 \times 0 \geq 0 \) shows adherence to this rule.
Even though \( S \) includes the zero vector, this alone is insufficient to define a subspace, emphasizing the importance of all subspace conditions working in tandem.
Vector Spaces
Understanding vector spaces is foundational for grasping subspaces. A vector space is a collection of objects called vectors, which can be added together or multiplied by scalars, obeying specific rules.
A subspace essentially inherits these properties and rules but applies them to a more limited set within the larger vector space.
  • Examples of vector spaces include \( \mathbb{R}^n \) where \( n \) represents the number of dimensions.
  • Each vector space must satisfy conditions like closure under addition and scalar multiplication, existence of a zero vector, and more.
Every potential subspace is expected to fulfill these rules to be considered legitimate.
However, if any requirements, like closure under addition or inclusion of the zero vector, are violated in any set proposal like our \( S \) in this exercise, it fails as a subspace.
Engaging with these concepts helps recognize how structured and diligent the requirements are for establishing vector spaces and subspaces, ensuring they perform predictably and consistently.

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Most popular questions from this chapter

(a) Prove that a graph is bipartite if and only if its vertices can be labeled so that its adjacency matrix can be partitioned as \\[ A=\left[\begin{array}{ll} O & B \\ B^{T} & O \end{array}\right] \\] (b) Using the result in part (a), prove that a bipartite graph has no circuits of odd length.

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